Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.

Divisibility Tests

Stage: 3, 4 and 5

Article by Tim Rowland

Published January 1997,October 2010,February 2011.

Here is another method for testing divisibility by $7$ suggested by
T.R.Mukundan

Subtract a multiple of $7$ from the number such that, on
subtraction, the final (units) digit of the remainder is zero.
(e.g. If the final digit of the given number is $7$, subtract $7$,
if the final digit is $5$ subtract $35$). This is always possible
since there is a unique number in the $7$ times table (up to $10
\times 7 = 70$) ending in each digit $0$ to $9$. (see note
below)

Ignore the zero on subtraction and repeat the above procedure
until only a number with two digits is obtained. If this number is
divisible by $7$, then the original number is also divisible by
$7$

Example:

Is $1778$ divisible by $7$?

The final digit of $1778$ is $8$, so we subtract $28$ ($4
\times 7$)

$$1778-28 = 1750$$

$1750$ is divisible by $7$ if and only if $175$ is, so we can
ignore the $0$. The final digit of $175$ is $5$, so we subtract
$35$ ($5 \times 7$)

$$175-35=140$$

$140$ is divisible by $7$ if and only if $14$ is, so we can
ignore the $0$. But $14=2\times 7$, so $1778$ is divisble by
$7$.

So really this is a method for dividing by $7$, rather than
just testing for divisibility. The divisibility test given in the
article was rather hard to remember, so maybe in this case we would
be better off just doing the division!

Note

In step 1 of the method, we used the fact that there is a unique
number in the $7$ times table (up to $10 \times 7 = 70$) ending in
each digit $0$ to $9$. Is this true in other times tables? It works
for $3, 7,9, 11\ldots$. But it certainly doesn't work for $2$ - no
multiple of $2$ ends in $1$. It turns out that works for any number
not divisible by $2$ or $5$ (the prime factors of $10$). To
understand why this is true, read this
article on the Chinese Remainder Theorem.

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